Question

Silver chromate, Ag.CrO, has a Kp of 9.00x 1012 Calculate the solubility in mol/L of silver chromate a) 1.31 x 10M d) 2.08 x 10 M c) 2.25 x 10-12M b) 1.65 x 10" M e) 1.50 x 106*M lsa for evothermic reactions and a value for

Answer 1

Answer:
`1.31* 10^4moles/L`

Step-by-step explanation:

Solubility product is defined as the equilibrium constant in which a solid ionic compound is dissolved to produce its ions in solution. It is represented as
`K_(sp)`

The equation for the ionization of the silver chromate is given as:

`Ag_2CrO_4\leftrightharpoons 2Ag^++CrO_4^(2-)`

We are given:

Solubility of
`Ag_2CrO_4` = S mol/L

By stoichiometry of the reaction:

1 mole of
`Ag_2CrO_4` gives 2 moles of
`Ag^(+)` and 1 mole of
`CrO_4^(2-)`.

When the solubility of
`Ag_2CrO_4` is S moles/liter, then the solubility of
`Ag^(+)` will be 2S moles\liter and solubility of
`CrO_4^(2-)` will be S moles/liter.

Expression for the equilibrium constant of
`Ag_2CrO_4` will be:

`K_(sp)=[Ag^+]^2[CrO_4^(2-)]`

`9.00* 10^(12)=[2s]^2[s]=4s^3`

`s=1.31* 10^4M`

Hence, the solubility of
`Ag_2CrO_4` is
`1.31* 10^4moles/L`